Base | Representation |
---|---|
bin | 111110100100110001111… |
… | …111111011000111110001 |
3 | 120020002022102212100110001 |
4 | 332210301333323013301 |
5 | 1030423101340312423 |
6 | 13051234510110001 |
7 | 622446341206603 |
oct | 76446177730761 |
9 | 16202272770401 |
10 | 4300101104113 |
11 | 1408731a73991 |
12 | 59547b1ab301 |
13 | 252661866559 |
14 | 10c1a9a7d173 |
15 | 76cc748acad |
hex | 3e931ffb1f1 |
4300101104113 has 2 divisors, whose sum is σ = 4300101104114. Its totient is φ = 4300101104112.
The previous prime is 4300101104111. The next prime is 4300101104149. The reversal of 4300101104113 is 3114011010034.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 3437082399969 + 863018704144 = 1853937^2 + 928988^2 .
It is a cyclic number.
It is not a de Polignac number, because 4300101104113 - 21 = 4300101104111 is a prime.
Together with 4300101104111, it forms a pair of twin primes.
It is not a weakly prime, because it can be changed into another prime (4300101104111) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2150050552056 + 2150050552057.
It is an arithmetic number, because the mean of its divisors is an integer number (2150050552057).
Almost surely, 24300101104113 is an apocalyptic number.
It is an amenable number.
4300101104113 is a deficient number, since it is larger than the sum of its proper divisors (1).
4300101104113 is an equidigital number, since it uses as much as digits as its factorization.
4300101104113 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 144, while the sum is 19.
Adding to 4300101104113 its reverse (3114011010034), we get a palindrome (7414112114147).
The spelling of 4300101104113 in words is "four trillion, three hundred billion, one hundred one million, one hundred four thousand, one hundred thirteen".
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